Question

For a quadratic function that models the height above ground of a projectile, how do you determine the maximum height or the time when the projectile reaches the ground?

Answer 1

Answer:

a. y(t) = - 1/2gt² + y₀ b. $t = \frac{v_{0} +/- \sqrt{v_{0} ^{2} + 2 y_{0}g } }{g }$

Step-by-step explanation:

Let the quadratic function y(t) = v₀t - 1/2gt² + y₀ represent the quadratic function that models the height above the ground of the projectile.

a. Maximum Height

At maximum height, the velocity, v₀ = 0, so substituting v₀ = 0 into the equation, we have

y(t) = v₀t - 1/2gt² + y₀

y(t) = 0 × t - 1/2gt² + y₀

y(t) = 0 - 1/2gt² + y₀

y(t) = - 1/2gt² + y₀

b. Time when the projectile reaches the ground

The time when the projectile reaches the ground is gotten when y(t) = 0, So

y(t) = v₀t - 1/2gt² + y₀

0 = v₀t - 1/2gt² + y₀

re-arranging, we have

1/2gt² - v₀t - y₀ = 0

Using the quadratic formula,

$t = \frac{-(-v_{0}) +/- \sqrt{(-v_{0}) ^{2} - 4 X (-y_{0}) X\frac{g}{2} } }{2 X \frac{g}{2} } \\= \frac{v_{0} +/- \sqrt{v_{0} ^{2} + 2 y_{0}g } }{g }$